Graded Rings

Graded Rings

If the index set \(I\) is a commutative monoid, we agreed to call a family of abelian groups \((A_i)_{i\in I}\) a graded abelian group. (§Abelian Groups, ⁋Definition 16) At the time, since there were no conditions on \(A_i\), this was not a particularly interesting definition; but now that a multiplication structure is added on \(A_i\), this definition becomes more meaningful.

Definition 1 Let a commutative monoid \(I\) and an \(I\)-indexed family of abelian groups \((A_i)_{i\in I}\) be given. If a multiplication structure defined on \(A=\bigoplus_{i\in I} A_i\) makes it into a ring, and additionally satisfies the condition

\[A_i A_j\subseteq A_{i+j}\qquad\text{for all $i,j\in I$}\]

then we call \(A\) an \(I\)-indexed graded ring. Elements belonging to \(A_i\) are called homogeneous elements.

By definition, any element of \(A\) can be uniquely expressed as a finite sum of homogeneous elements.

Proposition 2 If every element of \(I\) is cancellable and \(A=\bigoplus_{i\in I} A_i\) is a graded ring, then \(A_0\) is a subring of \(A\).

Proof

From \(A_0A_0\subseteq A_0\), \(A_0\) is closed under multiplication. Thus it suffices to show that in \(A=\bigoplus A_i\), the multiplicative identity \(1\) belongs to \(A_0\). Let \(1=\sum_{i\in I} \epsilon_i\). Then for any \(j\in I\) and any \(\alpha\in A_j\),

\[\alpha=1\alpha=\sum_{i\in I} \epsilon_i\alpha\in A_j\]

Now, for each \(i\), we have \(\epsilon_i\alpha\in A_{i+j}\), and since every element of \(I\) is cancellable, \(i+j=j\) holds only when \(i=0\). Therefore, comparing both sides of the above equality in each component of \(A=\bigoplus_{i\in I}A_i\), we see that \(\epsilon_i\alpha=0\) for all \(i\neq 0\), and \(\epsilon_0\alpha=\alpha\). Since every element of \(A\) can be expressed as a sum of homogeneous elements, \(\epsilon_0x=x\) for all \(x\in A\). Substituting \(x=1\) yields \(1=\epsilon_0\in A_0\).

In most cases, what we are interested in is the case where \(I=\mathbb{Z}\) or \(I= \mathbb{N}\). Therefore, the hypothesis of Proposition 2 is satisfied.

Example 3 For any abelian group \(G\), consider the free ring generated by \(G\), \(F(G)=\bigoplus_{n\geq 0} G^{\otimes n}\). Since the product of an element of \(G^{\otimes m}\) and an element of \(G^{\otimes n}\) belongs to \(G^{\otimes (m+n)}\), \(F(G)\) is an \(\mathbb{N}\)-graded ring. (§Definition of a Ring, §§Free Rings over Abelian Groups)

Graded Ring Homomorphisms

Definition 4 For a commutative monoid \(I\) and two \(I\)-graded rings \(A,A'\), a ring homomorphism \(\phi:A \rightarrow A'\) is a graded homomorphism if for every \(i\in I\), \(\phi(A_i)\subseteq A_i'\) holds.

It is not difficult to see that \(I\)-graded rings and \(I\)-graded homomorphisms form a category \(\bgr_I\Ring\).

Homogeneous Ideals and Quotients of Graded Rings

For a graded ring \(A=\bigoplus_{i\in I} A_i\) and a two-sided ideal of \(A\), \(\mathfrak{a}\), the quotient ring \(A/\mathfrak{a}\) may fail to inherit the grading of \(A\). That is, there may not exist a grading on \(A/\mathfrak{a}\) that makes the quotient map a graded homomorphism.

Example 5 Fix a commutative ring \(A\), and consider the polynomial ring whose coefficients are elements of \(A\),

\[A[\x]=\\{\alpha_n\x^n+\cdots+\alpha_1\x+\alpha_0\mid n\in\mathbb{N}, \alpha_i\in A\\}\]

Then this carries the structure of a graded ring via the decomposition

\[A[\x]=\bigoplus_{n\geq 0} A\x^n\]

On the other hand, consider the ideal generated by \(\x-1\), \((\x-1)\). Then as rings

\[A[\x]/(\x-1)\cong A\]

and explicitly, this isomorphism is obtained from the evaluation map defined by

\[\alpha_n\x^n +\cdots+\alpha_1\x+\alpha_0\quad \mapsto\quad \alpha_n+\cdots+\alpha_1+\alpha_0\]

by applying the first isomorphism theorem. However, in this case, the quotient map \(\pi:A[\x]\rightarrow A[\x]/(\x-1)\) sends, for any \(n\), \(A\x^n\) to all of \(A\). Therefore, if \(A\neq 0\), the sum of the \(\pi(A\x^n)\) cannot be a direct sum, and a grading that makes \(\pi\) a graded homomorphism does not exist on \(A[\x]/(\x-1)\).

To avoid this, we introduce the notion of a homogeneous ideal.

Proposition 6 For an \(I\)-graded ring \(A=\bigoplus_{i\in I} A_i\) and a two-sided ideal \(\mathfrak{a}\) of \(A\), the following are all equivalent.

1. \(\mathfrak{a}\) is the sum of the \(\mathfrak{a}\cap A_i\).
2. Whenever an arbitrary element of \(\mathfrak{a}\) is decomposed into homogeneous elements, each of these elements also belongs to \(\mathfrak{a}\).
3. \(\mathfrak{a}\) is generated by homogeneous elements.

Proof

As elements of \(A\), all elements of \(\mathfrak{a}\) can be uniquely written as a sum of homogeneous elements. Therefore, the first two conditions are equivalent to each other, and under condition 1, \(\mathfrak{a}\) is generated by the elements of \(\mathfrak{a}\cap A_i\), which are homogeneous elements, so condition 1 implies condition 3. Now we assume the third condition and prove the second. Suppose that \(\mathfrak{a}\) is generated by homogeneous elements \((x_j)_{j\in J}\). Then an arbitrary \(x\in \mathfrak{a}\) is written in the form

\[x=\sum_{k\in K} a_k x_{j(k)} b_k,\qquad\text{$K$ finite, $a_k,b_k\in A$}\]

Now each of \(a_k\) and \(b_k\), again as elements of \(A\), is written as a sum of homogeneous elements

\[a_k=\sum_{p} a_{kp},\qquad b_k=\sum_q b_{kq}\]

so we obtain

\[x=\sum_{k\in K}\sum_{p,q}a_{kp}x_{j(k)}b_{kq}\]

Here, each \(a_{kp}x_{j(k)}b_{kq}\) is a homogeneous element and all belong to \(\mathfrak{a}\). Now, grouping together terms of the same degree gives each homogeneous component of \(x\), and since these all belong to \(\mathfrak{a}\), we obtain condition 2.

A two-sided ideal satisfying the above equivalent conditions is called a homogeneous ideal. Then the following holds.

Proposition 7 For a homogeneous ideal \(\mathfrak{a}\), \(A/\mathfrak{a}\) is a graded ring, and its decomposition is given by the formula

\[A/\mathfrak{a}=\bigoplus_{i\in I}A_i/(\mathfrak{a}\cap A_i)\]

Proof

Let \(\pi:A\rightarrow A/\mathfrak{a}\) be the quotient map. Since \(A=\bigoplus_i A_i\) and \(\pi\) is surjective, \(A/\mathfrak{a}\) is the sum of the \(\pi(A_i)\), and because the kernel of the restriction of \(\pi\) to \(A_i\) is \(\mathfrak{a}\cap A_i\), we have \(\pi(A_i)\cong A_i/(\mathfrak{a}\cap A_i)\).

To show that this sum is a direct sum, we assume that a finitely supported family of homogeneous elements \((\alpha_i)\) satisfies \(\sum_i \pi(\alpha_i)=0\) and must show that \(\pi(\alpha_i)=0\) for all \(i\). By assumption, \(\sum_i\alpha_i\in\mathfrak{a}\), so by the second condition of Proposition 6 each \(\alpha_i\) must belong to \(\mathfrak{a}\), and hence to \(\mathfrak{a}\cap A_i\). That is, \(\pi(\alpha_i)=0\) for all \(i\).

Finally, from \(A_iA_j\subseteq A_{i+j}\) we obtain \(\pi(A_i)\pi(A_j)\subseteq \pi(A_{i+j})\), so this decomposition gives a graded ring structure on \(A/\mathfrak{a}\).

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References

[Bou] Bourbaki, N. Algebra I. Elements of Mathematics. Springer. 1998.